Combination of Capacitors question from JEE Main 2018 (15 Apr, Shift 1)

Answer: (a)

1. The two $5\ \mu\text{F}$ and the top $2\ \mu\text{F}$ all join the node after $6\ \mu\text{F}$ to the same lower node: $5 + 5 + 2 = 12\ \mu\text{F}$.
2. From that lower node to B: $4 \parallel 2 = 6\ \mu\text{F}$.
3. $12$ in series with $6$: $4\ \mu\text{F}$; in series with $6$: $\frac{6\times 4}{10} = 2.4\ \mu\text{F}$.
4. So the answer is (a) $2.4\ \mu\text{F}$.